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7 votes
1 answer
3k views

When does the determinant distribute over addition?

When does $\det(A+B)=\det(A)+\det(B)$ hold? I actually wonder if there is an easy answer for when $Per(A+B)=Per(A)+Per(B)$.
Turbo's user avatar
  • 13.9k
3 votes
1 answer
2k views

Eigenvalues of a block matrix with zero diagonal blocks

Suppose $A$ is a $k_1\times k_2$ matrix with real entries, $k_1<k_2$. Let $M$ be the matrix \begin{equation} M:=\begin{pmatrix} 0_{k_1} & A\\ A^\top & 0_{k_2} \end{pmatrix}, \end{equation} ...
AdamNie's user avatar
  • 53
3 votes
0 answers
122 views

Algebra of block matrices with scalar diagonals

I am interested in block matrices $A$, that is $A\in M_{n\times n}(R)$ where $R=M_{s\times s}(k)$ and $k$ is a field, such that for every positive integer $m$ the matrix $A^m$ has only scalar blocks ...
Adam Przeździecki's user avatar
3 votes
0 answers
1k views

Eigenvalues of block-hermitian matrices with zero diagonal blocks

I have a matrix of the form $$D = \left( \begin{array}{cc} 0 & C \\ C^{\dagger} & 0 \end{array} \right)$$ where $C$ is not necessarily hermitian. In general, can we say anything about the ...
Unwieldy Bob's user avatar
2 votes
1 answer
508 views

Jordan decomposition of a block matrix

Assume $A$ is a block matrix of the form: $$A=\left[\begin{array}{cccc} A_{11}&A_{12}&\ldots&A_{1n}\\ A_{21}&A_{22}&\ldots&A_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ ...
user293017's user avatar
2 votes
0 answers
172 views

Minimum of $\mathrm{rank}\left( \boldsymbol{W} \boldsymbol{H} \right)$, with $\boldsymbol{W}$ block diagonal

Let us assume that we have a full-rank $(n\cdot l)\times k$ matrix, $\boldsymbol{H}$, with no specific structure (e.g., a realization of a Gaussian i.i.d. random matrix), and an $m\times (n\cdot l)$ ...
Juan's user avatar
  • 61
2 votes
0 answers
535 views

eigenvalues of a square block matrix

How can we show that there are not defective eigenvalues for this square block matrix of dimension $2d \times 2d $: \begin{bmatrix} A&B\\-B& 0 \end{bmatrix} where A, B are real matrices, $A =\...
UserName's user avatar